Analysis of QPSK Satellite link and BER
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A satellite communication link transmits a QPSK signal at a rate of 10 Msps. The transmitter and receiver have ideal RRC filters with roll-off factor r = 0.25. At the input to the QPSK demodulator in the receiver, the CNR is 18.5 dB under clear air conditions (no rain in the link).
a. What is the bandwidth occupied by the QPSK signal?
b. What is the BER at the output of the receiver in clear air conditions, and the average time between errors? Give your answer in years, days, or hours as appropriate. Assume an implementation margin of 0.5 dB.
c. Rain in the satellite to earth path causes the overall CNR in the earth station receiver to fall to 15.5 dB. The link has an implementation margin of 0.5 dB.
What is the BER at the output of the receiver, and the average time between errors? Give your answer in hours, minutes, or seconds, as appropriate.
d. Why is it necessary to include an implementation margin when estimating the BER expected in a real digital communications link?
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Solution Summary
A satellite communication link transmits a QPSK signal at a rate of 10 Msps. The transmitter and receiver have ideal RRC filters with roll-off factor r = 0.25. At the input to the QPSK demodulator in the receiver, the CNR is 18.5 dB under clear air conditions (no rain in the link). Using these inputs the link is explored to determine
a. The bandwidth occupied by the QPSK signal?
b. The BER at the output of the receiver in clear air conditions, and the average time between errors
c. The BER at the output of the receiver given that the overall CNR in the earth station receiver falls to 15.5 dB under rain conditions. The link has an implementation margin of 0.5 dB.
The purpose of adding in modem implementation margin is then considered
Solution Preview
(a) The occupied bandwidth B is given in terms of the symbol rate SR by (1) below.
B = (1 + r)*SR (1)
Putting in values where r = 0.25, SR = 10 MSps we get
B = (1 + 0.25)*10 MHz
B = 12.5 MHz
(b) Received CNR(dB) is 18.5 dB.
The carrier level (C) is the product of the energy per bit Eb and the bit rate R given by (2).
C = Eb*R (2)
The noise power N can be expressed as the product of the noise spectral density No and the noise bandwidth (occupied bandwidth) B by (3).
N = No*B (3)
The CNR is simply the ratio (2)/(3), so we can say
CNR = Eb*R/No*B (4)
Re-arranging (4) we get an expression for the energy per bit to noise spectral density given by (5).
Eb/No = CNR*{B/R} (5)
Expressing (5) as a dB sum
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